2013-1434
DESCRIPTION
Unsaturated Hydrocarbons Alkenes. 2013-1434. Learning Objectives. Chapter two discusses the following topics and the student by the end of this chapter will: Know the structure, hybridization and bonding of alkenes Know the c ommon and IUPAC naming of alkenes - PowerPoint PPT PresentationTRANSCRIPT
Unsaturated HydrocarbonsAlkenes
1435-14362014-2015
Learning ObjectivesChapter two discusses the following topics and the student by the end of this chapter will:
Know the structure, hybridization and bonding of alkenes
Know the common and IUPAC naming of alkenes
Know the geometry of the double bond i.e. cis/trans isomerization
Know the physical properties of alkenes
Know the different methods used for preparation of alkenes (elimination reactions ; dehydrogenation, dehydration and alkenes stability (Zaitsev’s rule) play an important role in understanding these reactions
Know the addition reactions of alkenes and the effect of Markovnikov’s rule in determining the regioselectivity of this reaction.
Alkenes
They are unsaturated hydrocarbons – made up of C and H atoms and contain one or more C=C double bond somewhere in their structures.
Their general formula is CnH2n - for non-cyclic alkenes
Their general formula is CnH2n-2 - for cyclic alkenes
Structure Of Alkenes
3
Alkenes
Trigonal planar
2s2 2p3 3 x sp2 2p
sp2 Hybridization Of Orbitals In Alkenes
The electronic configuration of a carbon atom is 1s22s22p2
Thus
promotion hybridization
2p2 2s1
Alkenes
In ethylene (ethene), each carbon atom use an sp2 orbital to form a single
C-C bond. Because of the two sp2 orbitals overlap by end- to- end the resulting bond is called σ bond. The pi (π) bond between the two carbon atoms is formed by side- by-side overlap of the two unhybridized p- orbitals (2p–2p ) for maximum overlap and hence the strongest bond, the 2p orbitals are in line and perpendicular to the molecular plane. This gives rise to the planar arrangement around C=C
bonds. Also s orbitals of hydrogen atoms overlap with the sp2 orbitals in carbon atoms to form two
C-H bonds with each carbon atom. The resulting shape of ethene molecule is planar with
bond angles of 120º and C=C bond length is 1.34 Å
5
Orbital Overlap In Ethene
Alkenes
two sp2 orbitals overlap to form a sigma bond between the two carbon atoms
Orbital Overlap In Ethene
two 2p orbitals overlap to form a pi bond between the two carbon atoms
s orbitals in hydrogen atoms overlap with the sp2 orbitals in carbon atoms to form C-H bonds
the resulting shape is planar with bond angles of 120º and C=C (1.34 Å)
sp2 hybridized carbon atoms
6
Alkenes
Nomencalture Of Alkenes And Cycloalkenes
1. Alkene common names:
Substituent groups containing double bonds are: H2C=CH– Vinyl group H2C=CH–CH2– Allyl group
7
BrCl
Common: Allyl bromide Vinyl chlorride
H2C CH2 CH3-CH CH2 C CH2
CH3
H3C
Common: IsobuteneEthylene Propylene
Alkenes
2. IUPAC Nomenclature Of Alkenes
Find the longest continuous Carbon chain containing the double bond this determines the root name then add the suffix -ene.
Number the C- chain from the end that is nearer to the double bond. Indicate the location of the double bond by using the number of the first atom of the double bond just before the suffix ene or as a prefix.
Indicate the positions of the substituents using numbers of carbon atoms to which they are bonded and write their names in alphabetical order (N.B. discard the suffixes tert-, di, tri,---when alphabetize the substituents) and if more than one substituent of the same type are present use the prefixes di- or tri or tetra or penta,--- to indicate their numbers.
CH3CH CHCH2CH2CH3
Hex-2-ene or 2-Hexene(not 4-Hexene)
1 2 43 5 6
8
H2C CH CH2CH3
But-1-ene or 1-Butene (not 3-Butene)
1 2 3 4
Alkenes
CH3
CH3C CH
2-Methyl-but-2-eneor 2-Methyl-2-butene(not 3-Methyl-2-butene)
31CH3
2 4
CH3
Cl
3-Chloro-2-hexene(not 2-Chloro-1-methyl-1-pentene)
2
1
3
4
5
62
3
4
5
6
7
8
CH3-CH2-CH2-CH=CH-C-CH3
2,2-Dibromo-3-heptene(not 6,6-Dibromo-4-heptene)
1567 234(CH3CHCHCH2OCH3)
1-Methoxy-but-2-ene(not 4-Methoxy-but-2-ene)
OCH32
3 1
8
12
56
734
2,3,7-Trimethyl-non-3-ene(not 2-Isopropyl-6-methyl--2-octene)
9CN
4-Cyano-2-ethyl-1-pentene(not 2-Ethyl-4-cyano-1-pentene)
1
23
45
5-Methylcyclopenta-1,3-dieneAn ''a'' is added due to inclusion of di put two consonants consecutive
1
23
CH3
=
Br
Br
6-Methyl-2-octene
1
4
Alkenes
In cycloalkenes the double bond carbons are assigned ring locations #1 and #2. Which of the two is #1 may be determined by the nearest substituent rule.
If the substituents on both sides of the = bond are at the same distance, the numbering should start from the side that gives the substituents with lower alphabet the lower number.
10 3-tert-Butyl-7-isopropyl-cycloheptene(not 3-Isopropyl-7-tert-butylcycloheptene)
CH36 2
3
4
5
1-Methyl cyclopentene(not 2-Methylcyclopeneten)
CH3H3C
3,5-Dimethyl-cyclohexene(not 4,6-Dimethylcyclohexen)
(not 1,5-Dimethyl-2-cyclohexen)
1
1
Alkenes
When the longer chain cannot include the C=C, a substituent name is used.
6
2
3
5
1
1
CH CH2
Vinyl-cyclohexane
CH CH2
3-Vinyl-cyclohexene
4
Alkenes
Geometrical Isomerism In Alkenes G. I. found in some, but not all, alkenes It occurs in alkenes having two different groups / atoms attached to each carbon
atom of the = bond
G. I. x G. I. X G. I. G. I.
12
BA
DC
A=C or B=D No Cis or transe
(G. I. X)
A≠C B≠D, A =B or C=D Cis / A =D or C= B transe
G. I.
Alkenes
13
It occurs due to the Restricted Rotation of C=C bonds so the groups on either end of the bond are ‘fixed’ in one position in space; to flip between the two groups a bond must be broken.
X
Geometrical isomers can not convert to each at room temperature.
Alkenes
Geometrical Isomerism In Alkenes
A) Cis / trans isomerism in alkenes Exhibited by alkenes having two H’s and two other similar groups or
atoms attached to each carbon atom of the = bond (or generally the alkene have only two types of atom or groups i.e. ABC=CAB)
Cis prefix used when hydrogen atoms on both carbon atoms are on the SAME side of C=C bond
Trans prefix used when non-hydrogen groups / atoms are on the opposite sides of C=C bond
14
Types Of Geometric Isomerism
Cis-But-2-ene Trans-But-2-ene
Alkenes
CisGroups / atoms are on theSame Side of the double
bond
TransGroups / atoms are on
Opposite Sides across the double bond
15
H
ClCl
H H
ClH
Cl
cis-1,2-Dichloro-ethene trans-1,2-Dichloro-ethene
Trans-Oct-4-eneH H
Cis-Oct-4-ene
=
Alkenes
16
If the groups attached to the C=C are different, we distinguish the two isomers by adding the prefix Z (from German word Zusammen) if the higher-priority groups are together in the same side or E (from German word Entgegen) if the higher-priority groups are opposite sides depending on the atomic number of the atoms attached to each end of the C=C.
Atoms with higher atomic numbers receive higher priority I> Br > Cl > F > O > N > C > H
B) Z/ E isomerism in alkenes
Alkenes
17
CH3
BrI
ClBr
CH3I
Cl
CH
CH3
O
OH
CH3
I
Cl
H
CH2
Z-2-bromo-1-chloro-1-iodopropene E-2-bromo-1-chloro-1-iodopropene
Z Z
Alkenes
Exercise
Q1-Which of the following compounds can exhibit cis / trans isomerisma) 2-Methylpropeneb) 1-Butenec) 2-Methyl-2-pentened) 2-Butene e) 3-Methyl-2-hexeneQ2- Name the following compounds according to IUPAC system
18
c)b)a)
Alkenes
19
Br12
34
3-Bromo propene1
23
45
2-Ethyl-4-methyl pentene
CH3
Cl
CH3
1
23
4
5 6
4-Chloro-3,6-dimethylcyclohexene 3-Chloro-2,5-dimethylcyclohexene
CH3
Cl
CH3
6
12
3
4 5
C C
H
Br Br
H
C C
H
H Cl
Cl
Cis-1,2-DibromoetheneGeometrical isomerism
1,1-Dichloroethene
Trans-trans-2,4-heptadiene Trans-1,3,5-heptatriene
Cis-cis-2,5-heptadiene Home work
not geometrical isomerism
1
2
3
4
5
6
7
Alkenes
C C
H
H3C CH3
CH2CH3
1
2 3
4 5E-3-Methyl-2-pentene
Physical Properties of Alkenes
Alkenes are nonpolar compounds thus: Insoluble in water Soluble in nonpolar solvents ( hexane, benzene,…) The boiling point of alkenes increase as the number of carbons
increase.
20
Alkenes
Preparation Of Aalkenes1- Dehydration of alcohols ( removal of OH group and a proton from
two adjacent carbon atoms ) using mineral acids such as H2SO4 or H3PO4
CH3CH2OH H+/ heat
CH2 CH2 + OH2
OH
H+ OH2
H+/ heat
cyclohexanol cyclohexene
Ethanol Ethene
21
Alkenes
22
Zaitsev’sRule
If there are different protons can be eliminated with the hydroxyl group or with halogen atom, in this case more than one alkene can be formed, the major product will be the alkene with the most alkyl substituents attached to the double bonded carbon.
H3CCH3
OH
H / Heat
H2CCH3
H3CCH3
+ H2O
+ H2O
1- Butene Minor
2- Butene Major
Zaitsev rule: an elimination occurs to give the most stable, more highly substituted alkene
Alkenes
2- Dehydrohalogenation of alkyl halides using a base
23
Alkenes
or NaOH
24
3. Dehalogenation of vicinal dihalides
For example: Dehalogenation of 1,2-Dibromobutane leads to the formation of 1-Butene. In the presence of catalyst.
Br
Zn/AcOH
Br
Alkenes
Reactions Of Alkenes
25
Reactions of Alkenes
Oxidation Reactions
Addition(Electrophilic) reaction: - Hydrogenation - Halogenation - Hydrohalogenation - Halohydrin formation -Hydration
KMnO4
Ozonolysis
Alkenes
26
Alkenes
An electrophile, an electron-poor species,(from the Greek words meaning electron loving). It is a species (any molecule, ion or atom) that accept a pair of electrons to form a new covalent bond.
A nucleophile, an electron-rich species, ,(from the Greek words meaning nucleus loving). It is a species (any molecule, ion or atom) that donate an electron pair to form a new covalent bond.
C , H , Br , Cl , I , etc., AlCl3 , BF3 , FeCl3 , FeBr3 , etc.
C , OH , Br , Cl , I , etc., H2N, HS, etc., H2O ,
CH3OH , RNH2 , R2NH , R3N , , etc.
27
Electrophilic Addition Reaction1- Additions To The Carbon-Carbon Double Bond1.1 Addition Of Hydrogen: Hydrogenation
Alkenes
A
A
A
A+ H2
H H
A
A
A
A
An alkene An alkane
Pt or Ni or Pd
CH2 CH2 + H2Pt
CH3 CH3
CH3CH2
+ H2Pt CH3 CH3
CH3
CH3
+ H2Pt
CH3
CH3Cis-1,2-Dimethyl cyclohexane
1.2.Addition of Halogens( Halogenation)
28
Alkenes
A
A
A
A+ X2
X X
A
A
A
A (X= Cl or Br)
CH3CH3
+ Cl2CCl4 CH3 CH3
Cl
Cl
+ Br2CCl4
Br
Br
CH3
CH3
+ Br2CCl4
CH3
Br
Br
CH3
Trans-1,2- Dibromo-1,2-Dimethyl cyclohexane
1.3. Addition of Hydrogen Halides
29
However, if the double bond carbon atoms are not structurally equivalent, i.e. unsymmetrical alkenes as in molecules of 1- propene, 1-butene, 2-methyl-2-butene and 1-methylcyclohexene, the reagent may add in two different ways to give two isomeric products. This is shown for 1-propene in the following equation.
Only one product is possible from the addition of these strong acids to symmetrical alkenes such as ethene, 2-butene and cyclohexene.
A
AA
A+ HX
AA
H X
AA
(x= Cl or Br or I)
+ HClH3C
CH3
Cl
H
+ HIH
I
Alkenes
30
Alkenes
HBrCH3CHCH3
CH3CH3CH3
CH3CH=CH2
Br
Br
CH3CHCH3
CH3CH2CH2Br
Br2o Carbocation
1o Carbocation
maijor
minor
Stability of carbocation
CCH3H3C
CH3
CHH3C
CH3
CH2CH2 CH3
3o 2o 1o
31
Alkenes
Markovnikov’s rule stats that : In addition of unsymmetrical reagent to unsymmetrical alkenes the positive ion adds to the carbon of the alkene that bears the greater number of hydrogen atoms and the negative ion adds to the other carbon of the alkene.
However when the addition reactions to such unsymmetrical alkenes are carried out, it was found that 2-bromopropane is nearly the exclusive product. Thus it said the reaction proceeded according to Markovnikov’s rule
+ HClH3C
CH3
CH3
H3CCH3
CH3Cl
1.4. Addition of HOX halogen in aqueous solution ( -OH, X+): Halohydrin formation
32
Only one product is possible from the addition of HOX acids (formed from mixture of H2O and X2) to symmetrical alkenes such as ethene and cyclohexene.
A
AA
A+ H2O / X2
AA
OH X
AA (x= Cl or Br )
H3CCH3
Cl
OH
+ H2O / Cl2
Symmetrical akenes
Alkenes
33
Alkenes
However, addition reactions to unsymmetrical alkenes will result in the formation of Markovonikov’s product preferentially.
Cl
+ H2O / Cl2 OH
Unsymmetrical akenes
CH2Br
OH
+ H2O / Br2
1.5. Addition of H2O: Hydration
34
Only one product is possible from the addition of H2O in presence of acids as catalysts to symmetrical alkenes such as ethene and cyclohexene.
However, addition reactions to unsymmetrical alkenes will result in the formation of Markovonikov’s product preferentially.
CH3 CH3
OH
H
+ H2O
H
Unsymmetrical akenes
Symmetrical akenesA
AA
A+ H2O
AA
H OH
AA
H3CCH3
OH
H
+ H2O
H
H
Alkenes
1- Ozonolysis (Oxidative cleavage): This reaction involves rupture of the C=C to give aldehydes or ketones according to the structure of the original alkene.
35
A
AA
A+ O3
AA
O O
AA
O
Zn /H2O- H2O2
OA
A
+ OA
A( A= H or R)
i) O3
ii) Zn /H2O O + O
i) O3
ii) Zn /H2O O + O
H
i) O3
ii) Zn /H2O
O
O
Alkenes
Oxidation Reaction:
36
2- Oxidation with KMnO4 (Oxidative addition):
OH
OH
KMnO4 / OH
Alkenes
Cis- doil
Thank You for your kind attention !
Questions?Comments
37
Alkenes